/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 4 A quiz question gives random sam... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A quiz question gives random samples of n=10observations from each of two Normally distributed populations. Tom uses a table of t distribution critical values and 9degrees of freedom to calculate a 95%confidence interval for the difference in the two population means. Janelle uses her calculator’s two-sample t interval with 16.87degrees of freedom to compute the 95%confidence interval. Assume that both students calculate the intervals correctly. Which of the following is true?

(a) Tom’s confidence interval is wider.

(b) Janelle’s confidence interval is wider.

(c) Both confidence intervals are the same.

(d) There is insufficient information to determine which confidence interval is wider.

(e) Janelle made a mistake; degrees of freedom have to be a whole number.

Short Answer

Expert verified

The true statement is option (a) Tom’s confidence interval is wider.

Step by step solution

01

Concept introduction

The normal distribution, also known as the Gaussian distribution, is a symmetric probability distribution centered on the mean, indicating that data around the mean occur more frequently than data far from it.

02

Explanation

Janelle, on the other hand, employs more degrees of freedom than Tom.

A lower t*-value is associated with a higher degree of freedom.

A lower t*-value means a smaller margin of error and, as a result, a smaller confidence interval.

Then we know that Janelle's confidence interval is narrower than Tom's confidence interval, or Tom's confidence interval is wide than Janelle's confidence interval.

Thus the answer (a) is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a baseball game, 42of 65randomly selected people report owning an iPod. At a rock concert occurring at the same time across town, 34of 52randomly selected people to report owning an iPod. A researcher wants to test the claim that the proportion of iPod owners at the two venues is different. A 90%confidence interval for the difference in population proportions is ( 0.154, 0.138). Which of the following gives the correct outcome of the researcher’s test of the claim?

(a) Since the confidence interval includes 0, the researcher can conclude that the proportion of iPod owners at the two venues is the same.

(b) Since the confidence interval includes 0, the researcher can conclude that the proportion of iPod owners at the two venues is different.

(c) Since the confidence interval includes 0, the researcher cannot conclude that the proportion of iPod owners at the two venues is different.

(d) Since the confidence interval includes more negative than positive values, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game.

(e) The researcher cannot draw a conclusion about a claim without performing a significance test.

Did the treatment have an effect? The investigators expected the control group to adjust their breeding date the next year, whereas the well-fed supplemented group had no reason to change. The report continues: "But in the following year, food-supplemented females were more out of synchrony with the caterpillar peak than the controls." Here are the data (days behind caterpillar peak):

Carry out an appropriate test and show that it leads to the quoted conclusion.

The U.S. Department of Agriculture (USDA) conducted a survey to estimate the average price of wheat in July and in September of the same year. Independent random samples of wheat producers were selected for each of the two months. Here are summary statistics on the reported price of wheat from the selected producers, in dollars per bushel:

Construct and interpret a 99% confidence interval for the difference in the mean wheat price in July and in September.

40. Household size How do the numbers of people living in households in the United Kingdom (U.K.) and South Africa compare? To help answer this question, we used Census At School’s random data selector to choose independent samples of 50 students from each country. Here is a Fathom dotplot of the household sizes reported by the students in the survey.

Refer exercise 35.Suppose we select independent SRSs of 25men aged 20to 34and 36boys aged 14and calculate the sample mean heights xMand xB.

(a) Describe the shape, center, and spread of the sampling distribution ofxM-xB.

(b) Find the probability of getting a difference in sample means xM-xBthat’s less than0mg/dl. Show your work.

(c) Should we be surprised if the sample mean cholesterol level for the 14-year-old boys exceeds the sample mean cholesterol level for the men? Explain.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.